document.write( "Question 977624: p(x)=8x^2+6x+2, c=1/2 Use synthetic division and the remainder theorem to find
\n" ); document.write( "
\n" ); document.write( "p(c) which in this case is p(1/2) \n" ); document.write( "
Algebra.Com's Answer #599175 by Edwin McCravy(20055) $\"\"$ $\"About$
You can put this solution on YOUR website!
\r\n" );
document.write( "The remainder theorem says (in words):\r\n" );
document.write( "\r\n" );
document.write( "If you have a polynomial function p(x) = (some terms in powers of x)\r\n" );
document.write( "Then if you want to find p(c), then\r\n" );
document.write( "\r\n" );
document.write( "Instead of substituting c for x in p(x), you can get the same answer\r\n" );
document.write( "by performing synthetic division with the polynomial with c on the\r\n" );
document.write( "far left of the synthetic divison, and taking ONLY the REMAINDER.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We put c = ½ to the left of the synthetic division   \r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "  ½ | 8  6  2\r\n" );
document.write( "    |________\r\n" );
document.write( "\r\n" );
document.write( "Bring down the 8 below the bottom line:\r\n" );
document.write( "\r\n" );
document.write( "  ½ | 8  6  2\r\n" );
document.write( "    |________\r\n" );
document.write( "      8\r\n" );
document.write( "\r\n" );
document.write( "Multiply the 8 times the ½ getting (8)(½) = 4\r\n" );
document.write( "and write 4 above the line and to the right of the 8,\r\n" );
document.write( "underneath the 6:\r\n" );
document.write( "\r\n" );
document.write( "  ½ | 8  6  2\r\n" );
document.write( "    |    4    \r\n" );
document.write( "      8\r\n" );
document.write( "\r\n" );
document.write( "Add 6+4 getting 10, and write 10 below to line underneath\r\n" );
document.write( "the 4 to the right of the 8:\r\n" );
document.write( "\r\n" );
document.write( "  ½ | 8  6  2\r\n" );
document.write( "    |    4   \r\n" );
document.write( "      8 10\r\n" );
document.write( "\r\n" );
document.write( "Multiply the 10 times the 0.5 getting (10)(½) = 5\r\n" );
document.write( "and write 5 above the line and to the right of the 10,\r\n" );
document.write( "underneath the 2, beside the 4:\r\n" );
document.write( "\r\n" );
document.write( "  ½ | 8  6  2\r\n" );
document.write( "    |    4  5\r\n" );
document.write( "      8 10 \r\n" );
document.write( "\r\n" );
document.write( "Add 2+5 getting 7, and write 7 below to line underneath\r\n" );
document.write( "the 5 to the right of the 10:\r\n" );
document.write( "\r\n" );
document.write( "  ½ | 8  6  2\r\n" );
document.write( "    |    4  5\r\n" );
document.write( "      8 10  7\r\n" );
document.write( "\r\n" );
document.write( "So p(½) = the last number on the bottom is the remainder, \r\n" );
document.write( "which is 7. and that's equal to p(½)\r\n" );
document.write( "\r\n" );
document.write( "Check by substituting ½ for x in \r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So you see that you get the same answer 7 when you do synthetic\r\n" );
document.write( "division with ½ on the far left as you get when you substitute ½\r\n" );
document.write( "for x in the polynomial.  It's usually easier to use the remainder\r\n" );
document.write( "theorem than to substitute -- especially when the polynomial has \r\n" );
document.write( "lots of terms.  This one only has 3 terms so the remainder theorem\r\n" );
document.write( "is not that time-saving on this one, but it is when the polynomial\r\n" );
document.write( "has more terms.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );